A) \[\frac{m\nu r}{{{I}_{0}}+m{{r}^{2}}}\]
B) \[\frac{2m\nu r}{{{I}_{0}}}\]
C) \[\frac{\nu }{2r}\]
D) \[\frac{m\nu r}{2{{I}_{0}}}\]
Correct Answer: A
Solution :
Given that \[{{I}_{0}}\] is the moment of inertia of table and gun and m the mass of bullet. Initial angular momentum of system about centre \[{{\operatorname{L}}_{i}} =\left( {{I}_{0}} + m{{r}^{2}} \right){{\omega }_{0}}\] Let co be the angular velocity of table after the bullet is fired. Final angular momentum \[{{L}_{f}}={{I}_{0}}\omega -\,\,m(\nu - r\omega )r\] where \[(\nu -r\omega )\] is absolute velocity of bullet to the right. Equating (i) and (ii), we get \[(\omega -{{\omega }_{0}})=\frac{m\nu r}{{{I}_{0}}+m{{r}^{2}}}\]You need to login to perform this action.
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